Analyzing Equation $(\cos\beta - 1)x^2 + \cos\beta x + \sin\beta = 0$ for Real Roots

We need to find the condition under which the equation $(\cos\beta - 1)x^2 + (\cos\beta)x + \sin\beta = 0$ has a real root for $x$, given that $\beta \in [0, \pi]$.

Let's represent the equation in the standard quadratic form $Ax^2 + Bx + C = 0$. Here:

• $A = \cos\beta - 1$
• $B = \cos\beta$
• $C = \sin\beta$

The existence of real roots depends on whether the equation is linear ($A=0$) or quadratic ($A \neq 0$) and the value of its discriminant ($\Delta = B^2 - 4AC$).

Linear Case Analysis ($A = 0$)

The equation becomes linear if the coefficient of $x^2$ is zero. $A = \cos\beta - 1 = 0$ This implies $\cos\beta = 1$. For the given range $\beta \in [0, \pi]$, this condition is met only when $\beta = 0$.

Let's substitute $\beta = 0$ into the original equation:

• $A = \cos 0 - 1 = 1 - 1 = 0$
• $B = \cos 0 = 1$
• $C = \sin 0 = 0$

The equation simplifies to $0 \cdot x^2 + 1 \cdot x + 0 = 0$, which means $x = 0$. This is a real root.

So, the condition $1 - \cos\beta = 0$ allows for a real root.

Quadratic Case Analysis ($A < 0$)

For any $\beta$ in the interval $[0, \pi]$, we know that $\cos\beta \le 1$. Therefore, the coefficient $A = \cos\beta - 1$ is always less than or equal to zero ($A \le 0$).

We already analyzed the case $A=0$. Now let's consider $A < 0$, which occurs when $\cos\beta < 1$. For $\beta \in [0, \pi]$, this corresponds to $\beta \in (0, \pi]$.

When $A < 0$, the equation is a quadratic equation. It has real roots if its discriminant is non-negative ($\Delta \ge 0$).

Calculating the discriminant:

$ \Delta = B^2 - 4AC $

Substitute the values of A, B, and C:

$ \Delta = (\cos\beta)^2 - 4(\cos\beta - 1)(\sin\beta) $

$ \Delta = \cos^2\beta - 4\cos\beta \sin\beta + 4\sin\beta $

Let's rearrange the terms:

$ \Delta = \cos^2\beta + 4\sin\beta(1 - \cos\beta) $

Now, we analyze the sign of $\Delta$ for $\beta \in (0, \pi]$:

• The term $\cos^2\beta$ is always greater than or equal to 0.
• In the interval $(0, \pi]$, $\sin\beta$ is strictly positive ($\sin\beta > 0$).
• Also, for $\beta \in (0, \pi]$, $\cos\beta < 1$, meaning $1 - \cos\beta$ is strictly positive ($1 - \cos\beta > 0$).

Since $\sin\beta > 0$ and $(1 - \cos\beta) > 0$, the term $4\sin\beta(1 - \cos\beta)$ is strictly positive.

Therefore, $\Delta$ is the sum of a non-negative term ($\cos^2\beta$) and a positive term ($4\sin\beta(1 - \cos\beta)$):

$ \Delta = \cos^2\beta + 4\sin\beta(1 - \cos\beta) \ge 0 + (\text{positive value}) $

This confirms that $\Delta > 0$ for all $\beta \in (0, \pi]$. So, the quadratic equation always has real roots in this case.

Overall Condition for Real Roots

We have established that:

• A real root exists when $1 - \cos\beta = 0$.
• Real roots exist when $1 - \cos\beta > 0$.

Combining both scenarios, real roots exist for all values of $\beta$ in the interval $[0, \pi]$. The condition that covers all these possibilities is $1 - \cos\beta \ge 0$.

This condition ensures that the coefficient $A$ is non-positive and, when $A<0$, the discriminant is positive, guaranteeing at least one real root ($x$) for the given equation.